find the critical numbers of the function. (enter your answers as a comma - separated list.)\n\n$g(t)=t\\sqrt…

Explanation:

Step1: Find the derivative of (g(t))

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = t) and (v=\sqrt{16 - t}=(16 - t)^{\frac{1}{2}}). (u^\prime=1), (v^\prime=\frac{1}{2}(16 - t)^{-\frac{1}{2}}\times(- 1)=-\frac{1}{2\sqrt{16 - t}}). So (g^\prime(t)=\sqrt{16 - t}+t\times\left(-\frac{1}{2\sqrt{16 - t}}\right)=\frac{2(16 - t)-t}{2\sqrt{16 - t}}=\frac{32-2t - t}{2\sqrt{16 - t}}=\frac{32 - 3t}{2\sqrt{16 - t}}).

Step2: Set (g^\prime(t) = 0) and find (t)

Set the numerator equal to zero: (32-3t = 0), then (3t=32), so (t=\frac{32}{3}). Also, consider where the derivative is undefined. The derivative (g^\prime(t)) is undefined when (16 - t=0) (denominator is zero), i.e., (t = 16). But since (t<15), we discard (t = 16).

Answer:

(\frac{32}{3})